Question

if ABCD is a cyclic quadrilateral show that Cos A + Cos B + cosC+Cos D =0​

Answer 1

Opposite angles in a cyclic quadrilateral are supplementary.

• A + C = 180°
• B + D = 180°

Consider A + C = 180°.

$\sf \: A + C = 180 \\ \\ \implies \sf \: A = 180 - C$

Taking cosine on both sides,

$\implies \sf \: cosA = cos(180 - C) \\ \\ \implies \sf \: cosA = cos(- C)$

If cosine has a negative argument, the resulting value is equal to negative of the same value of cos with a positive argument.

$\implies \sf \: cosA = - cosC$

Similarly,

$\implies \sf \: cos B = - cos D$

Now,

$\sf \: cos A + cos B + cosC+cos D = cos A + cos B - cos A - cos B \\ \\ \implies \boxed{ \boxed{ \sf cos A + cos B + cosC+cos D = 0}}$

Hence, proved.

Answer 2

Answer:

As A, B, C, D are the vertices of CyQuad ABCD,

A+C=180° and B+D=180°

By the rules of *associated angles*, we can write,

CosC=Cos(2x90-A)= -CosA..(1)

We can apply the same process on CosB or CosD..

Then we shall get,

CosB=-CosD.. (2)

So, in L. H. S.

CosA+CosB+CosC+CosD

=CosA-CosD-CosA+CosD.. From 1&2

=0

Hence, L.H.S=R.H.S., |PROVED|

Explanation:

thanks..